Verify the statements for Riemann-integrable function $f_n(x)$.

Question

For each $n = 1, 2, \cdots$ a function $f_n(x)$ is defined so that it is Riemann-integrable on $[a, b]$ and the series $\sum_{n=1}^{\infty}f_n(x)$ converges $\forall \space x \in [a,b]$.

Which of the following statements are true? $$\lim_{n\rightarrow\infty}\sup_{x\in[a,b]}|f_n(x)| = 0$$ $$\int_{a}^{b}\bigg(\lim_{n\rightarrow \infty} |f_n(x)|\bigg) dx = 0$$ $$\lim_{n\rightarrow\infty}\int_{a}^{b}|f_n(x)|dx = 0$$

ATTEMPT

I. $f_n$ is integrable, so it is bounded. Therefore, $\sup |f_n(x)|$ exists and equals to $|f_n(c)|, c\in [a, b]$. Since the series converges, $\lim f_n(x) = 0 \space \forall x \in [a, b]$. Hence $\lim \sup _{x \in [a, b]}|f_n(x)| =\lim|f_n(c)| = |0| = 0.$ Answers indicate that this is false, but I don't see where I am mistaken.

II. Recall that $\lim f_n(x) = 0 \space \forall x \in [a, b]$. Then $\lim |f_n(x)| = |0| = 0 \space \forall x \in [a, b]$. Therefore, the integrand is 0 on $[a, b]$, and so is the integral.

III. No idea.

It would be great if there was a counterexample.

Answer 1

Let $ f_n$ defined at $ [0,1]$ by $$f_n(x)=x^n \; if \; x\ne 1 \; and \; f_n(1)=0$$

$$f_n \; is \; Riemann\; integrable\; at \; [0,1],$$

$$\sum f_n(x) \; converges \; \forall x\in[0,1]$$

but

$$\sup_{x\in [0,1]}|f_n(x)|=1$$ the first statement is false.

Answer 2

As for the validity of the statements, I and III are both false. The example hamam-Abdallah gave, shows I is false. i.e,

$$f_n(x)=\begin{cases} x^n & ;0\leq x<1 \\ 0 &; x=1 \end{cases} $$

A similar example can show that III is false. Notice that for the $f_n$ given in said example, we have

$$ \int_0^1 f_n(x)dx=\int_0^1 x^ndx= \frac{x^{n+1}}{n+1}\vert_{x=0}^{x=1}=\frac{1}{n+1}. $$

So if you define

$$ g_n(x):=\begin{cases} (n+1)x^n &; 0\leq x<1 \\ 0 &; x=1 \end{cases}, $$

$\int_0^1 g_n(x)dx=1$, while $\sum_{n=1}^{\infty}g_n(x)$ converges for all $x\in [0,1]$.

I would just add as a side-note, that both I and III were true, were you to assume further that $f_n$ uniformly converges to the function $f\equiv0$ in $[a,b]$.